Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 25

Description

Greatest Common Divisor Traversal

You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nums[i], nums[j]) > 1, where gcd is the greatest common divisor.

Your task is to determine if for every pair of indices i and j in nums, where i < j, there exists a sequence of traversals that can take us from i to j.

Return true if it is possible to traverse between all such pairs of indices, or false otherwise.

Example 1:

Input: nums = [2,3,6]
Output: true
Explanation: In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.

Example 2:

Input: nums = [3,9,5]
Output: false
Explanation: No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.

Example 3:

Input: nums = [4,3,12,8]
Output: true
Explanation: There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
struct UnionSet {
  vector<int> parent;
public:
  UnionSet(int size): parent(size) {
    iota(parent.begin(), parent.end(), 0);
  }

  int find(int x) {
    if(parent[x] != x) parent[x] = find(parent[x]);
    return parent[x];
  }

  void merge(int x, int y) {
    x = find(x);
    y = find(y);
    parent[x] = y;
  }
};
class Solution {
public:
  bool canTraverseAllPairs(vector<int>& nums) {
    int mmax = *max_element(nums.begin(), nums.end());
    if(mmax == 1) return nums.size() == 1;
    UnionSet us(mmax + 1);
    vector<bool> visited(mmax + 1);
    for(auto n : nums) {
      for(int i = 2; i * i <= n; ++i) {
        if(n % i) continue;
        us.merge(n, i);
        us.merge(n, n / i);
      }
    }
    for(auto n : nums) {
      if(us.find(n) != us.find(mmax)) return false;
    }
    return true;
  }
};

// Accepted
// 925/925 cases passed (311 ms)
// Your runtime beats 80.84 % of cpp submissions
// Your memory usage beats 91.62 % of cpp submissions (86.9 MB)